\(\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 25, antiderivative size = 107 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=-\frac {\arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}-\frac {\text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}} \] Output:

-arctan(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a/f/(a*sin 
(f*x+e))^(1/2)-arctanh(cos(f*x+e)^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^( 
1/2)/a/f/(a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=-\frac {b \left (\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right )+\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \] Input:

Integrate[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(3/2),x]
 

Output:

-((b*(ArcTan[(Cos[e + f*x]^2)^(1/4)] + ArcTanh[(Cos[e + f*x]^2)^(1/4)])*Sq 
rt[a*Sin[e + f*x]])/(a^2*f*(Cos[e + f*x]^2)^(1/4)*Sqrt[b*Tan[e + f*x]]))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3081, 27, 3042, 3045, 266, 756, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}}dx}{\sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}}dx}{a \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \sin (e+f x)}dx}{a \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos (e+f x)}{a f \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{1-\cos ^2(e+f x)}d\sqrt {\cos (e+f x)}}{a f \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \int \frac {1}{\cos (e+f x)+1}d\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (e+f x)}d\sqrt {\cos (e+f x)}+\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )\right )}{a f \sqrt {a \sin (e+f x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \arctan \left (\sqrt {\cos (e+f x)}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (e+f x)}\right )\right )}{a f \sqrt {a \sin (e+f x)}}\)

Input:

Int[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(3/2),x]
 

Output:

(-2*(ArcTan[Sqrt[Cos[e + f*x]]]/2 + ArcTanh[Sqrt[Cos[e + f*x]]]/2)*Sqrt[Co 
s[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a*f*Sqrt[a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3081

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ 
n)   Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 
1)]) || IntegersQ[m - 1/2, n - 1/2])
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53

method	result	size
default	\(\frac {\left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\right )-\ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right )+2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {b \tan \left (f x +e \right )}\, \cos \left (f x +e \right )}{2 f \left (1+\cos \left (f x +e \right )\right ) a \sqrt {a \sin \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(164\)

Input:

int((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/f*(arctan(1/2/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2))-ln(2*(2*cos(f*x+e) 
*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/ 
2)-cos(f*x+e)+1)/(1+cos(f*x+e))))*(b*tan(f*x+e))^(1/2)*cos(f*x+e)/(1+cos(f 
*x+e))/a/(a*sin(f*x+e))^(1/2)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (91) = 182\).

Time = 0.26 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.86 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {-\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \left (f x + e\right )^{3} + 4 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, b \cos \left (f x + e\right )^{2} - 5 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, a f}, \frac {2 \, \sqrt {\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) - b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {\frac {b}{a}} \log \left (\frac {4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{a}} - {\left (b \cos \left (f x + e\right )^{2} + 6 \, b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right )}{4 \, a f}\right ] \] Input:

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

[1/4*(2*sqrt(-b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f 
*x + e))*sqrt(-b/a)*cos(f*x + e)/((b*cos(f*x + e) + b)*sin(f*x + e))) + sq 
rt(-b/a)*log(-(b*cos(f*x + e)^3 + 4*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + 
e)/cos(f*x + e))*sqrt(-b/a)*cos(f*x + e)*sin(f*x + e) - 5*b*cos(f*x + e)^2 
 - 5*b*cos(f*x + e) + b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + 
e) + 1)))/(a*f), 1/4*(2*sqrt(b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin 
(f*x + e)/cos(f*x + e))*sqrt(b/a)*cos(f*x + e)/((b*cos(f*x + e) - b)*sin(f 
*x + e))) + sqrt(b/a)*log((4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f* 
x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/a) - (b*cos(f*x + e)^2 + 
6*b*cos(f*x + e) + b)*sin(f*x + e))/((cos(f*x + e)^2 - 2*cos(f*x + e) + 1) 
*sin(f*x + e))))/(a*f)]
 

Sympy [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {b \tan {\left (e + f x \right )}}}{\left (a \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((b*tan(f*x+e))**(1/2)/(a*sin(f*x+e))**(3/2),x)
 

Output:

Integral(sqrt(b*tan(e + f*x))/(a*sin(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*tan(f*x + e))/(a*sin(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*tan(f*x + e))/(a*sin(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((b*tan(e + f*x))^(1/2)/(a*sin(e + f*x))^(3/2),x)
 

Output:

int((b*tan(e + f*x))^(1/2)/(a*sin(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}}{\sin \left (f x +e \right )^{2}}d x \right )}{a^{2}} \] Input:

int((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x)
 

Output:

(sqrt(b)*sqrt(a)*int((sqrt(tan(e + f*x))*sqrt(sin(e + f*x)))/sin(e + f*x)* 
*2,x))/a**2